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Unformatted text preview: — 31
Chapter 3 Problem Solutions 3.1 Determine the highest speed at which two generators mounted on the same
shaft can be driven so that the frequency of one generator is 60 Hz and the
frequency of the other is 25 Hz. How many poles does each machine have? Solution: Let P = number of poles: 2x60x60 2x60x25 speed = P60 = P25
{’29 _ E9
P25 ' 25 P60 and P25 must be even integral numbers lowest value where P50 = 2.41325. Thus, P25 = 10 P50 = 24 3.2 The threephase synchronous generator described in Example 3.1 is operated
at 3600 rpm and supplies a unity poWer factor load. If the terminal voltage of
the machine is 22 kV and the ﬁeld current is 2500 A, determine the line current
and the total power consumptiOn of the load. Solution: Using the values in the solution of Example 3.1, 45855 _
at“ = 3838 x2000V _ 29869.1 V
Given:
VLL = 22 kV
yam = (J2/J3’)x22000V=17962.9v If on = 17962.9 coswt, then in = in“: 0056.115 and ea! = 17962.9coswt —4.1484 x 10‘3 x 1207: x iamsinwt
17962.9 coswt — 1.563911.“ sin wt 17962.92 +(1.56391‘%“)2 = 29869.1 V
15259.4 A iam/ﬂ = 10.79 kA
x/Ex 22 x 10.79 x 1 MW = 411.2 MW m
D
II Hence, z"l a“
L:
P3¢> 3.3 A threephase roundrotor synchronous generator has negligible armature re
sistance and a synchronous reactance Xd of 1.65 per unit. The machine is connected directly to an inﬁnite bus of voltage 104$ per unit. Find the in
ternal voltage E. of the machine when it delivers a current of (a) 1.04 30° per 32 unit, (b) 1.04m per unit and (c) 1.04—30" per unit to the inﬁnite bus. Draw
phasor diagrams depicting the operation of the machine in each case. Solution:
Ei/Lo = VaLgi+IaL6XdL_QOO
= 1.04Q+1.0L9><1.65490°
= 1.0&°+1.65590°+0
(a)
0 = 30°
EiLé = 1.0&°+1.654120°
1.44183° per unit
Va=1.0p.u.
(b)
0 = 0°
= 1.0ﬂ+1.65490° E‘L5
t _ o . 1.65 p.u.
= 1.934 58.8 per umt 6 = 30°
Ei[_6 1.0&+1.65460°
= 2.324 38° per unit 33 3.4 A threephase roundrotor synchronous generator, rated 10 kV, 50 MVA has
armature resistance R of 0.1 per unit and synchronous reactance Xd of 1.65 per
unit. The machine operates on a 10 kV inﬁnite bus delivering 2000 A at 0.9
power factor leading. (a) Determine the internal voltage Bi and the power angle 6 of the machine.
Draw a phasor diagram depicting its operation. (b) What is the opencircuit voltage of the machine at the same level of exci
tation? (c) What is the steady—state shortcircuit current at the same level of excita
tion? Neglect all saturation effects. Solution: (a) Choosing Vb = 10 kV and MVAb = 50 MVA: Z, = (0.1+ j1.65) per unit = 1.653(8653" 2 244g
0
V = l—O per unit = 1.0 per unit
50 x 103
I = —— A = 2886.75A
b «E x 10
2000 . .
Ia. — 2886.75 per unlt — 0.693 per unit
9 = 005'1 0.9 = 258"
Big — VzL°+IaZaza+0
1.0L0'1l 0.693 x 1.6530; 112.37” per unit
= 1.195; 61.83° = 11.9516183‘j kV
(b) Opencircuit voltage: E, = 11.95 kV (c) Shortcircuit voltage: E1 1.195 E 1.653 per unit = 0.7242 per unit = 2090.7 A in; "‘2“ ~ 1 «a .;; «E ,5! n w 3%»: :gx‘.~a’s'zz‘ 125%”:3‘; V_;r:':;:'_;::; f1" .’ 15:: g ; 2 3.13:“:52: 1; a: :51; "r W: .gr 44:“ a v 5:} ta 4%. v gel—3% ‘ a—Js—sz: Esq—m: sail .» "Fa2,. .. ca: 445“.» 34 3.5 A threephase roundrotor synchronous generator, rated 16 kV and 200 MVA,
has negligible losses and synchronous reactance of 1.65 per unit. It is operated
on an inﬁnite bus having a voltage of 15 kV. The internal emf E, and the
power angle 6 of the machine are found to be 24 kV (lineto—line) and 274°,
respectively. ((1) Determine the line current and the threephase real and reactive power .
being delivered to the system. (b) If the mechanical power input and the ﬁeld current of the generator are
now changed so that the line current of the machine is reduced by 25% at the power factor of (a), ﬁnd the new internal emf E. and the power angle
6. (c) While delivering the reduced line current of (b), the mechanical power
input and the excitation are further adjusted so that the machine operates at unity power factor at its terminals. Calculate the new values of Bi and
6. Solution: (a) Using 16 kV, 200 MVA base; Va = 15/16 per unit = 09375 per unit _
EiLé = %4 274° per unit = 1.5127.4° per unit
EiLé—Vaﬂ = InXd490°—0
1.54 274° — 0.937540: = L, x 1.65; 90° — 9 In 4 90° — 0 = 0.4818160.27° per unit
In 50 = 0481842973" per unit
3
Base I = M kA = 7.217 kA
\/§ x 16 Therefore, I a
S 0.4818 x 7.217 kA = 3.477 kA
0.9375 x 0.4818 per unit = 0.4517 per unit
90.34 MVA
Thus, P = 90.34 cos 29.73° MW = 78.45 MW
Q = 90.34 sin 29.73° Mvar = 44.80 Mvar ll (5) New Ia. = 0.75 x 0.4818 per unit = 0.3614 per unit
90 — 6 = 60.27°
EiLb‘ = VGLO°+IQX4490°—6 0.9375¢°+ 0.3614 x 1.65 4 6027" = 1.3374 228° per unit
21.45228“ kV LL 35 NewIa = 0.3614 per unit 9 = 0°
Eiéé = 0.937540:+ 0.3614 x 1.654 90° 1.1114 325° per unit = 17.8; 325° kV .w
2:.
§ arm.“ . < 3.6 The threephase synchronous generator of Prob. 3.5 is operated on an inﬁnite
bus of voltage 15 kV and delivers 100 MVA at 0.8 power factor lagging. (a) Determine the internal voltage Ei, power angle 6 and the line current of
the machine. 3, (b) If the ﬁeld current of the machine is reduced by 10%, while the mechanical
power input to the machine is maintained constant, determine the new
value of 6 and the reactive power delivered to the system. (c) The prime mover power is next adjusted without changing the excitation
so that the machine delivers zero reactive power to the system. Determine
the new power angle 6 and the real power being delivered to the system. ., was. 1%; «H w ”again A; (d) What is the maximum reactive power that the machine can deliver if the
level of excitation is maintained as in (b) and (c)? Draw a phasor diagram for the operation of the machine in cases (a), (b) and (c). :
Solution: ‘ . (a) From Prob. 3.5, Va(or V,) = 0.9375 per unit
S = 0.5 per unit Xd = 1.65 per unit 0 = 36.9°
IO = S/Va = 0.5/0.9375 per unit
E./_6 = VGL01+IEX¢¢90°+0
0.5
= 0.9375 0° . ' ° — . ° '
L + 0.9375 x 1 6o 5 90 36 9 per unit 1.6258125]o per unit = 26.0; 25.7" kV 36 New E, 0.9 x 1.6258 per unit = 1.46322 per unit
P = 0.5 x 0.8 per unit 04 per unit . PXd ‘ . _ 0.4x 1.65
— _1 — = 1 ——————— = o
6 " 5m (1413.) 51“ (0.9375x1.46322) 2876
V X—d— ——(E c056— Vt)
0.9375 II New Q (1.46322 cos 2876" —— 0.9375)
0.196 per unit = 39.2 Mvar (c) When Q = 0, V 0.9375
— —1 —l = _1 = — o
6 —— cos E1 cos (1.46322) 00.15 EV 0.9375 1.463 2
P = —XT sin6= —1—:35—2 sin 5015" per unit = 0.638 per unit 127.65 MW (d) For V}, E. and Xd ﬁxed, Qmu occurs when 6 = 0°. Hence, V, 0.9375
ma = E' —V = . —0.9 7— '
Q 2 Xd ( t) 1.65 (1 46322 3 a) per umt
= 0.2787 per unit = 59.74 Mvar =1.626 p.u. v.=o.9375 p.11. 3.7 Starting with Eq. (3.31), modify Eq. (3.38) to show that lVl P = W{IEI(Rcosé+dein6)— thR}
V Q = R.“'X2{Xd<IEcos6Iv.1—RE.sm5} when the synchronous generator has nonzero armature resistance R. 37_ Solution:
From Eq. (365), V = E + Ia(R +de) and, therefore, lEilLé—thldﬁ Ia = .
(R'l'JXd) ,
I. : IEild—thl
“ (RJ'Xd) 3
. lellEild— W ‘1
S = P+ = VI. = ————.—— i
’Q ’ “ (12—an
= MHEzHcoséjsinél W
(RJ'Xd) Z
_ lellEiHcosé—jsiMMRi—J’Xd) — W (212+an l
_ (R2 +X3) :
Separating real and imaginary parts, 1
thllEx‘l . M121? ;
P (R2 +X3){ cos +dem } (R2+Xg) :
V
= Ez—li—lfg {IEil (Rcos6 + deiné) — [th R}
d
_ thllEil _ . Mlsz
Q _. (R2 +X3) {Xdcosé Rsm6} ——(R2 +X3)
lel m {X4 (IEilCOS5  thl) — R lEilSm 5} 3.8 The threephase synchronous generator described in Example 3.4 is now oper—
ated on a 25.2 kV inﬁnite bus. It is found that the internal voltage magnitude
lEil = 49.5 kV and that the power angle 6 = 38.5". Using the loading capa
bility diagram of Fig. 3.14, determine graphically the real and reactive power
delivered to the system by the machine. Verify your answers using Eqs. (3.38). Solution:
25.2
V, = 25.2 IN = W per unit = 1.05 per unit
49.5 . ‘ .
lEi[ = 49.5 kV = 3— per unit = 2.0625 per unit
The distance 72—]: on the loading capability diagram is
IEil 2.0625 “4le = ——1‘05 x 1.7241 units = 1.1393 units The angle formed by points kno is 38.5". Hence, point I: is marked as shown. By reading
from the chart, Pk = 0.7 per unit and QC = 0.31 per unit. P+jQ = IthQIPk+ij1 = 1.052(0.7+j0.31) per unit
= 1.052 (0.7+ 30.31) x 635 MVA
P = 490 MW Q = 217 Mvar 38 From Eq. (3.38), _ mm.
P — Xd sxné
1.05x2.0625 , c, _ ,
= mess ) x635MW  496.5 MW
V
Q = LX§<IE.icos6—Ivzl)
1.05 (2.0625cos(38.5°) — 1.05) x 635 Mvar = 218.2 Mvar
1.7241 3.9 A threephase salientpole synchronous generator with negligible armature re
sistance has the following values for the inductance parameters speciﬁed in Table 3.1, ,
L3 = 2.7656 mH Mf = 31.6950 mH Ln = 0.3771 mH
M3 = 1.3828 mH Lf; = 433.6569 mH During balanced steady—state operation the ﬁeld current and aphase armature
current of the machine have the respective values if = 4000 A in = 20, 000 sin (0d — 30°) A (a) Using Eq. (3.41), determine the instantaneous values of the flux linkages
Aa, Ab, Ac and A; when 6.1 = 60°. (b) Using Park’s Transformation given by Eqs. (3.42) and (3.43), determine the
instantaneous values of the flux linkages Ad, Ag and A0, and the currents id,
2}, and 2'0 when 1% = 60°. (c) Verify results using Eqs. (3.45)  (3.46) Solution: (11) From Table 3.1, Lao Lab Lac La _Ms —M_,
Labc é Lac Lab Lac = _Ms L: —M_,
Lac Lab Lac “My ‘Ms L: ‘
c0520.; —cos2(0¢ +30°) —cos2(9d + 150°)
+ L... —cos2(9d+30°) cos2(0d—120°) —cos2(0d—90°)
—cos2(0d+ 150°) cos2(9d —90°) cos2(6d+ 120°)
2.7656 —1.3828 l.3828
Lag,c = —1.3828 2.7656 —1.3828
1.3828 —1.3828 2.7656 cos 120° — cos 180° — cos 420°
+ 0.3771 — cos 180° cos (—120°) — cos (—60°) mH
— cos 420° — cos (—60°) cos 360° — '39 —1.0057 2.57705 l.57135
l.57135 —1.57135 3.1427 2.57705 1.0057 —1.57135
mH La; cos 8d
Labc,f é [ Lbf J = Mf COS (9d — 120°)
LC, cos (6,, — 240°)
cos 60° 15.8475
= 31.695 cos (—60°) mH = 15.8475 mH
cos (—_180°) ~31.695 lll> i,z sin (30°) 10
2'an 1;, = 20000 sin (—90°) A = 20 RA
ic sin (—210°) 10 With 2', = 4 kA and L f f = 433.6569 mH, )‘a Lao Lab Lac Laf in
Ab _ Lba Lbb Lbc Lbf i6
Ac — Lea ch Lac ch 1‘:
A! Lfa Lfb Lfc Ln if
2.57705 —1.0057 —1.57135 15.8475 10
_ —1.0057 2.57705 —1.57135 15.8475 —20 Wb—T
_ —1.57135 1.57135 3.1427 —31.6950 10
15.8475 15.8475 —31.6950 433.6569 4
93.5610
13.9215
= —79.6395 WM“
12592026 (b) When ed = 60°, 1 1 _1 1 1_ 1/3
2 2 2 7: ¢§ 3
= _ 6/3 _ﬁ = 1 _1
P 3 2 2 0 75 72' 0
L 1 1 1 1 1
v5 7? 75‘ 3 75 75
1 1 2
Ad 73 73 1/; 93.5610 97.5381
= 1 1 =
A., 75 —75 0 .13.9215 76.0016 WbT
1 1 1
A0 75 75 75 —79.6395 0
W
P A«be
 1 1 2 '
u 73 75 1/; 10 —12.2474
 = 1 _ 1 =
zq 75 75 0 —20 21.2132 kA
' 1 1 1
1,0 75 E 7—5 10 0
P “W
1nab: 40 L, = L3+M5—;Lm = 3.58275 mH
@111, = 38.8183 mH
‘ A4 = Ldid+¢§Mm = 4.71405 x (—12.2474)+38.8183 x4 WbT
= 97.5381 WbT
Ag = Lqiq = 3.58275x21.2132 WbT = 76.0016 WbT
A0 = Loio = 0 (since 10:0)
1, = ﬁMfid+me = 38.8183 x (—12.2474) +433.6569 x 4 war 1259.20 WbT 3.10 The armature of a three—phase salientpole generator carries the currents
z", = «5 x 1000 sin (81 — 0a) A
2'2, = \/§ x 1000 sin (6d — 120° — ea) A «5 x 1000 sin (0.: — 240° — an) A a.
n
l (0) Using the PTransformation matrix of Eq. (3.42), ﬁnd the directaxis cur
rent id and the quadratureaxis current iq. What is the zero.sequence cur—
rent 1'0? (b) Suppose that the armature currents are 1;, = 1/2' x1000sin(0d—0a) A
2'1, = ic = 0 Determine id, 1", and 1'0. Solution: (a) id 20
1}, = P ‘Lb
to re
1 1 .
= 3 15: 1c: (1) x mom/5 sfn (9d — 9a) 8
3 2 “2 5m(9a—0a—12O) A
712. 715 715 sin (0d — 6,, — 240°) H g
E
Q.
:1
Q)
n. I
Q:
0
+
W
o O
v t» 41 1 1d 2 § 5 —% sin(0d — 90)
 1. 1 .1. 0
10 J2 E «2'
é—sin (6d — 9a)
2000
\/§ 2 5111(041 a) i sin (9d — 05) 3.11 Calculate the directaxis synchronous reactance Xd, the directaxis transient
reactance X [1 and the directaxis subtransient reactance X {1’ of the 60 Hz' salient
pole synchronous machine with the following parameters: L5 = 2.7656 mH Lff = 433.6569 mH Ln = 4.2898 mH
M5 = 1.3828 mH Mf = 31.6950 mH MD = 3.1523 mH Ln 2 0.3771 mH M, = 37.0281 mH Solution:
3 , 3 Ld = L + Ms — 5L," = 2.7606+ 1.3828 — 5 x 0.3771 mH = 4.71405 mH
X1 = 1207.— x 4.71405 x 10—39 = 1.777 n , 3 M} _ 3 31.69502
Ld — Ld — E f}? — 4.71400—5 X 433.6569 mH — 1.2393 mH
x; = 12077 x 1.2393 x 10—30 = 0.467 n
L” L 3 MfLD'l'MIQDLff—szMDM, d ‘ “'5 LffLD—ME _ 4 71405 _3 316952 x 4.2898 + 3.15232 x 433.6569 — 2 x 31.6950 x 3.1523 x 37.0281
' 2 433.6569 x 4.2898 — 37.02812
0.9748 mH X}; = 12077 x 0.9748 x 1039 = 0.367 n 3.12 The single~1ine diagram of an unloaded power system is shown in Fig. 3.22.
Reactances of the two sections of transmission line are shown on the diagram.
The generators and transformers are rated as follows: ) 1nH . .mwmci‘mﬁu‘nm': :; 42 Generator 1: 20 MVA, 13.8 IN, ’1; = 0.20 per unit Generator 2: 30 MVA, 18 kV, Xé’ = 0.20 per unit Generator 3: 30 MVA, 20 kV, X (’1’ = 0.20 per unit Transformer T1: 25 MVA, 220Y/13.8A kV, X = 10% Transformer T 2: Singlephase units each rated 10 MVA,127/ 18 kV, X = 10%
Transformer T3: 35 MVA, 220Y/22Y kV, X = 10% ((1) Draw the impedance diagram with all reactances marked in per unit and
with letters to indicate points corresponding to the singleline diagram.
Choose a base of 50 MVA, 13.8 kV in the circuit of Generator 1. (b) Suppose that the system is unloaded and that the voltage throughout the
system is 1.0 per unit on bases chosen in part (a). If a threephase short cir—
cuit occurs from bus C to ground, ﬁnd the phasor value of the shortcircuit
current (in amperes) if each generator is represented by its subtransient
reactance. (c) Find the megavoltamperes supplied by each generator under the conditions
of part (b). Solution: (a) 50.20 j0.0826 C j0.1033 j0.167 Gen 1: X” = 0.2 x 553 = 0.50 per unit
391) rating T2 = 220/ 18 kV, 30 MVA
Base in trans. line: 220 kV, 50 MVA
Base for Gen 2 = 18 kV
Gen 2: X” = 0.2 x g = 0.333 per unit
Base for Gen 3 = 22 kV
20 2 50
Ge 1 H = . — — = . .
n3 X 02(22) x 30 0275perun1t
50 .
Transformer T1: X = .01 x g = 0.20 per unit
Transformer T2: X = .01 x 35% = 0.167 per unit
Transformer T3: X = .01 x 3—: = 0.143 per unit Transmission lines: 2
Base Z = 2?) = 968 Q
00
80 100
_ = . ' — = 0. O '
968 0 0826 per umt 968 1 33 per umt SIC 0.7826 per unit
0.6033 per unit
= 0.418 per unit ll per unit = 1.2785—90° per unit 1.658 5—90° per unit 7 = 0—41—85—900 per unit = 2.392(—90° per unit
.7 3 . ' Eh = E, = E,‘ = 1.0[_0: per unit
X1 —— 0.50 + 0.20 + 0.0826 per unit
X2 — 0.333 + 0.167 + 0.1033 per unit
X3 = 0.143 + 0.275 per unit
E 1 c,
11 ’ 3J7; ’ 0.78261399—
I  E1  1 —90° er unit
2 _ jXQ _ 0.6033; F
13 = E 1
If = I1 + I2 + I3 =
50 x 106
IﬁeatC = —————A = 131.22A
b J27 x 220 x 103 .
If = 5.328 x 131.22 A = 699 A 1521 = Eing = 1.0x 1.658 x 50 MVA = 82.9 MVA
{53] = EisIg = 1.0 x 2.392 X 50 MVA = 119.6 MVA 3.13 The ratings of the generators, motors and transformers of Fig. 3.23 are: Generator 1: Generator 2: Synchronous motor 3:
Threephase Y—Y transformers:
Threephase Y—A transformers: 20 MVA, 18 IN, 1;: 20% 20 MVA, 18 kV, X” = 20% 30 MVA, 13.8 kV, X” = 20% 20 MVA, 138Y/20Y kV, X = 10%
15 MVA, 138Y/13.8A kV, X = 10% 43 (1.278 + 1.658 + 2.392) (—90° per unit = 5.328 (490° per unit 44 (0) Draw the impedance diagram for the power system. Mark impedances in
per unit. Neglect resistance and use a base of 50 MVA, 138 kV in the 409
line. v (1)) Suppose that the system is unloaded and that the voltage throughout the
system is 1.0 per unit on bases chosen in part (a). If a threephase short cir
cuit occurs from bus C to ground, ﬁnd the phasor value of the shortcircuit
current (in amperes) if each generator is represented by its subtransient
reactance. (c) Find the megavoltamperes supplied by each synchronous machine under
the conditions of part (b). Solution: (0) Base voltages are: 40 {2 lines 138 kV
20 (2 lines 138 kV
Gen. 1 & 2 20 kV
Motor 3 13.8 kV . . . 1382
Base 1mpedance in lines = 30— = 381 Q
40
40 Q 1' : = — = . 0 ‘
me Z 381 0 1 5 per unit
. . _ 20 _ .
20 $2 hne. Z _. EST — 0.053 per umt
Transformers:
50
YY = 0.1 x 50 = 0.250 per unit
, 50 .
XA = 0.1 x ~1—5 = 0.333 per umt
Gens. 1 & 2: 2
18 50
XII = 0 _ — = 7 ~
20 x (20) x 20 0.40:: per unit 45 Motor 3: 50
X" = 0.20 x E = 0.333 per unit (b) If a fault occurs at C, by symmetry equal currents are input from generators 1 and 2.
Moreover, no current should exist between busses A and B through the 30.105 per unit
branch. If this branch is omitted from the circuit, the system simpliﬁes to S/C Eh  Eu = Eu = LOLOO. per unit
X1 — X2 = 0.405+O.250 +0053 + 0.333 per unit = 1.041 per unit
X3 — 0.333 per unit
_ _ [Eill __ 1.0 . _ .
[II] _ [12] _ 1X1} —. 1.041 per unit — 0.9606 per unit
Bi 1.0 . .
IIgI = I]X;: = 0.333 per unit = 3.0 per unit
[Ifl = 4.9212 per unit
50 x 106
135 tC = —————A=2091.8A
baa ﬁx13.8x103
w = 4.9212 x 2091.8 A = 10.29 kA
(C)
3511 = ngl = 1.0 x 50 x 0.9606 MVA = 48.03 MVA
33 = 1.0 x 50 x 3.0 MVA = 150 MVA Chapter 4 Problem Solutions 4.1 The allalumin conductor identiﬁed by the code word Bluebell is composed
of 37 strands ,ch having a diameter of 0.1672 in. Tables of characteristics
of all—aluminu conductors list an area of 1,033,500 cmil for this conductor (1 cmil = (7/4) x 10‘6 in2). Are these values consistent with each other? Find
the overall are of the strands in square millimeters. Solution: diameter = 0.1672x 1000 = 167.2 mils/strand ...
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 Spring '09
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